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I'm trying to write a function with an ifstream& argument.

void word_transform(ifstream & infile)
    infile("content.txt");
    //etc
which gave me an error:
  
Type 'ifstream' (aka 'basic_ifstream ') does not provide a call operator.
Can you please me what's wrong?
                Functions that take stream arguments usually work with a stream that is already open. For example, ifstream s("content.txt"); word_transform(s);.
– molbdnilo
                Dec 16, 2014 at 9:23
call operator is a function like operator()( params ) allowing to use the syntax myObject( params ).
So, when you write infile(...), you are trying to us a call operator.
What you are trying to do is to open a file, use the open method:
void word_transform(ifstream & infile)
    infile.open("content.txt",std::ios_base::in);
    if ( infile.is_open() )
        infile << "hello";
    infile.close();
But, as commented, it does not really make sense to pass infile reference to such a function. You may consider:
void word_transform(istream& infile)
    infile << "hello";
int main()
    ifstream infile;
    infile.open("content.txt",std::ios_base::in);
    if ( infile.is_open() )
        word_transform( infile );
    infile.close();
    return 0;
void word_transform()
    ifstream infile;
    infile.open("content.txt",std::ios_base::in);
    if ( infile.is_open() )
        infile << "hello";
    infile.close();
int main()
    word_transform();
    return 0;
                Your last comment is very pertinent: if he is going to open and close the file in the function, the std::ifstream should be local to the function; it serves no purpose to have it as an argument.  A more frequent idiom, however, would be to have it opened and closed by the caller, and to pass a std::istream& argument (rather than an std::ifstream&).
– James Kanze
                Dec 16, 2014 at 10:44
You attempt to call operator() on your parameter. That will not work. Are you trying to open a file? If you get an ifstream as parameter, it should be open from the start because you opened it outside your function. Passing a stream and then opening it inside your function does not make sense.
void word_transform(std::ifstream& infile)
    // read something from infile
int main()
   std::ifstream file("content.txt");
   // error checks
   word_transform(file);
   return 0;
Note that this would try to call operator() on already created object of type infile. As no such operator exists from ifstream , you got an error.
Rather you should do:-
infile.open("content.txt");
Typically, (references to) streams are passed to functions for I/O. This means the function should take
std::istream or std::ostream argument, but not std::ifstream or std::ofstream. This way your function can be used with any stream object derived from std::istream, including std::cin, std::istrstream, and std::ifstream.
As nvoigt said, passing an std::ifstream to a function for opening makes little sense. It is definitely not clear to the caller that its stream may be closed and opened to another file.
        
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